Is there something like 'nonlocal' in Python < 3?

Question

I got a piece of code like this:

foo = None

def outer():
    global foo
    foo = 0

    def make_id():
        global foo
        foo += 1
        return foo


    id1 = make_id() # id = 1
    id2 = make_id() # id = 2
    id3 = make_id() # ...

I find it ugly to have foo defined in the outermost scop, I would prefer to have it only in outer function. As I understand correctly, in Python3 this is done by nonlocal . Is there a better method for what I want to have? I would prefer to declare and assign foo in outer and maybe to delcare it global in inner :

def outer():
    foo = 0

    def make_id():
        global foo
        foo += 1     # (A)
        return foo

    id1 = make_id() # id = 1
    id2 = make_id() # id = 2
    id3 = make_id() # ...

(A) does not work, foo seems to be searched in the outermost scope.

Answer 1

I use 1-element lists for this purpose:

def outer():
    foo = [0]
    def make_id():
        r = foo[0]
        foo[0] += 1
        return r
    return make_id

make_id = outer()
id1 = make_id()
id2 = make_id()
...

This is the same as using nonlocal , at the cost of a slightly more cumbersome syntax ( foo[0] instead of foo ).

Answer 2

No, have this as a parameter to your make_id function. Even better put your id in a class, and have make_id as an instance method, and have that instance as a global (if you must).

Answer 3

不，最好的替代方法是函数属性。